Total sum except adjacent of a given node in BST
Given a BST and a key Node, find the total sum in BST, except those Node which are adjacent to key Node.
Examples:
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1:-First find the total sum of BST
2:-Search the key Node and trace its parent Node.
3:-If the key Node is present then, subtract the sum of its adjacent Node from total sum
4:-If key is not present in BST then return -1.
C++
// C++ program to find total sum except a given Node in BST#include <bits/stdc++.h>using namespace std;struct Node { int data; struct Node *left, *right;};// insertion of Node in TreeNode* getNode(int n){ struct Node* root = new Node; root->data = n; root->left = NULL; root->right = NULL; return root;}// total sum of bstint sum(struct Node* root){ if (root == NULL) return 0; return root->data + sum(root->left) + sum(root->right);}// sum of all element except those which are adjecent to key Nodeint adjSum(Node* root, int key){ int parent = root->data; while (root != NULL) { if (key < root->data) { parent = root->data; root = root->left; } else if (root->data == key) // key Node matches { // if the left Node and right Node of key is // not null then add all adjecent Node and // subtract from totalSum if (root->left != NULL && root->right != NULL) return (parent + root->left->data + root->right->data); // if key is leaf if (root->left == NULL && root->right == NULL) return parent; // If only left child is null if (root->left == NULL) return (parent + root->right->data); // If only right child is NULL if (root->right == NULL) return (parent + root->left->data); } else { parent = root->data; root = root->right; } } return 0;}int findTotalExceptKey(Node *root, int key){ return sum(root) - adjSum(root, key);}// Driver codeint main(){ struct Node* root = getNode(15); root->left = getNode(13); root->left->left = getNode(12); root->left->left->left = getNode(11); root->left->right = getNode(14); root->right = getNode(20); root->right->left = getNode(18); root->right->right = getNode(24); root->right->right->left = getNode(23); root->right->right->right = getNode(25); int key = 20; printf("%d ", findTotalExceptKey(root, key)); return 0;} |
Java
// Java program to find total sum// except a given Node in BSTclass GFG{static class Node{ int data; Node left, right;};// insertion of Node in Treestatic Node getNode(int n){ Node root = new Node(); root.data = n; root.left = null; root.right = null; return root;}// total sum of bststatic int sum(Node root){ if (root == null) return 0; return root.data + sum(root.left) + sum(root.right);}// sum of all element except those// which are adjecent to key Nodestatic int adjSum(Node root, int key){ int parent = root.data; while (root != null) { if (key < root.data) { parent = root.data; root = root.left; } else if (root.data == key) // key Node matches { // if the left Node and right Node of key is // not null then add all adjecent Node and // subtract from totalSum if (root.left != null && root.right != null) return (parent + root.left.data + root.right.data); // if key is leaf if (root.left == null && root.right == null) return parent; // If only left child is null if (root.left == null) return (parent + root.right.data); // If only right child is null if (root.right == null) return (parent + root.left.data); } else { parent = root.data; root = root.right; } } return 0;}static int findTotalExceptKey(Node root, int key){ return sum(root) - adjSum(root, key);}// Driver codepublic static void main(String[] args){ Node root = getNode(15); root.left = getNode(13); root.left.left = getNode(12); root.left.left.left = getNode(11); root.left.right = getNode(14); root.right = getNode(20); root.right.left = getNode(18); root.right.right = getNode(24); root.right.right.left = getNode(23); root.right.right.right = getNode(25); int key = 20; System.out.printf("%d ", findTotalExceptKey(root, key));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to find total sum# except a given Node in BSTclass getNode: def __init__(self, n): self.data = n self.left = None self.right = None# Total sum of bstdef sum(root): if (root == None): return 0 return (root.data + sum(root.left) + sum(root.right))# Sum of all element except those# which are adjecent to key Nodedef adjSum(root, key): parent = root.data while (root != None): if (key < root.data): parent = root.data root = root.left elif (root.data == key): # Key Node matches # if the left Node and right Node of key is # not None then add all adjecent Node and # subtract from totalSum if (root.left != None and root.right != None): return (parent + root.left.data + root.right.data) # If key is leaf if (root.left == None and root.right == None): return parent # If only left child is None if (root.left == None): return (parent + root.right.data) # If only right child is None if (root.right == None): return (parent + root.left.data) else: parent = root.data root = root.right return 0def findTotalExceptKey(root, key): return sum(root) - adjSum(root, key)# Driver codeif __name__ == '__main__': root = getNode(15) root.left = getNode(13) root.left.left = getNode(12) root.left.left.left = getNode(11) root.left.right = getNode(14) root.right = getNode(20) root.right.left = getNode(18) root.right.right = getNode(24) root.right.right.left = getNode(23) root.right.right.right = getNode(25) key = 20 print(findTotalExceptKey(root, key)) # This code is contributed by bgangwar59 |
C#
// C# program to find total sum// except a given Node in BSTusing System;class GFG{class Node{ public int data; public Node left, right;};// insertion of Node in Treestatic Node getNode(int n){ Node root = new Node(); root.data = n; root.left = null; root.right = null; return root;}// total sum of bststatic int sum(Node root){ if (root == null) return 0; return root.data + sum(root.left) + sum(root.right);}// sum of all element except those// which are adjecent to key Nodestatic int adjSum(Node root, int key){ int parent = root.data; while (root != null) { if (key < root.data) { parent = root.data; root = root.left; } else if (root.data == key) // key Node matches { // if the left Node and right Node of key is // not null then add all adjecent Node and // subtract from totalSum if (root.left != null && root.right != null) return (parent + root.left.data + root.right.data); // if key is leaf if (root.left == null && root.right == null) return parent; // If only left child is null if (root.left == null) return (parent + root.right.data); // If only right child is null if (root.right == null) return (parent + root.left.data); } else { parent = root.data; root = root.right; } } return 0;}static int findTotalExceptKey(Node root, int key){ return sum(root) - adjSum(root, key);}// Driver codepublic static void Main(String[] args){ Node root = getNode(15); root.left = getNode(13); root.left.left = getNode(12); root.left.left.left = getNode(11); root.left.right = getNode(14); root.right = getNode(20); root.right.left = getNode(18); root.right.right = getNode(24); root.right.right.left = getNode(23); root.right.right.right = getNode(25); int key = 20; Console.Write("{0} ", findTotalExceptKey(root, key));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to find total sum// except a given Node in BSTclass Node{ constructor(data) { this.left = null; this.right = null; this.data = data; }};// Insertion of Node in Treefunction getNode(n){ let root = new Node(n); return root;}// Total sum of bstfunction sum(root){ if (root == null) return 0; return root.data + sum(root.left) + sum(root.right);}// Sum of all element except those// which are adjecent to key Nodefunction adjSum(root, key){ let parent = root.data; while (root != null) { if (key < root.data) { parent = root.data; root = root.left; } // key Node matches else if (root.data == key) { // If the left Node and right Node of key is // not null then add all adjecent Node and // subtract from totalSum if (root.left != null && root.right != null) return (parent + root.left.data + root.right.data); // If key is leaf if (root.left == null && root.right == null) return parent; // If only left child is null if (root.left == null) return (parent + root.right.data); // If only right child is null if (root.right == null) return (parent + root.left.data); } else { parent = root.data; root = root.right; } } return 0;}function findTotalExceptKey(root, key){ return sum(root) - adjSum(root, key);}// Driver codelet root = getNode(15);root.left = getNode(13);root.left.left = getNode(12);root.left.left.left = getNode(11);root.left.right = getNode(14);root.right = getNode(20);root.right.left = getNode(18);root.right.right = getNode(24);root.right.right.left = getNode(23);root.right.right.right = getNode(25);let key = 20;document.write(findTotalExceptKey(root, key));// This code is contributed by divyeshrabadiya07</script> |
Output:
118
Time Complexity: O(n) + O(h) where n is number of nodes in BST and h is height of BST. We can write time complexity as O(n).
